A complex number $w\in D$ is a fixed point for the map $f:D \to D$ if $f(w)=w$.
Prove that if $f:D\to D$ is analytic and has two distinct fixed points, then $f$ is the identity,that is,$f(z)=z$ for all $z\in D$.
If $f(0)=0$ , I can use Schwarz lemma to show that $f(z)=ze^{i\theta}$ and $\theta=0$. How could I deal with the condition when $f(0)=z_0 \neq 0$.
Best Answer
Proof assuming that $D$ is the open unit disk: Let $z_1,z_2$ be two distinct fixed points of $f$. Let $g(z)=\frac {z-z_1}{1-\overset {-}z_1 z}$. Let $h=g\circ f \circ g^{-1}$. Then $h$ masp $D$ into itself, vanishes at $0$ and it has a second fixed point, namely $g(z_2)$. By your argument based on Schwarz lemma you get $h=$ identity. This implies $f=$ identity. [I have used the fact that functions of the form $\frac {z-a}{1-\overset {-}a z}$ (where $|a| <1)$ are bijective, bi-holomorphic maps of the open unit disk].